1. Introduction - University of Michiganners312/CourseLibrary/Lecture01.pdf · 1Aside: Later on, when we discuss relativistic kinematics, we shall see that Q = (m1 +m2 −m a −m

1. Introduction

1.1. About this course ...According to the College Bulletin:

NERS 311. Elements of Nuclear Engineering and Radiological Sciences I

Photons, electrons, neutrons, and protons. Particle and wave properties of radiation.

Introduction to quantum mechanics. Properties and structure of atoms.

What is it really about? It’s about Modern Physics.See: http://en.wikipedia.org/wiki/Modern physics

Modern Physics covers just about everything in physics from the 1900s onward, andexplains those things that Classical Physics could not.

Elements of Nuclear Engineering and Radiological Sciences I NERS 311: Slide #1

1.2 Modern Physics encompasses ...

Quantum Mechanics: The study of really small, light things, like neutrons, protons,nuclei, atoms, and slow, non-relativistic electrons ... the basic building blocks ofNuclear Engineering. We spend most of our time in 311 doing this.

Relativistic Quantum Mechanics: The study of very fast (approaching c), light things.More cool stuff, but not covered in detail for NERS. We do make use of its results.

General Relativity: The study of very fast (approaching c) objects, with external forces.Also, the study of very dense, massive objects, like black holes. Cool stuff, but not

covered for NERS.


Modern Physics encompasses (continued)...

Special Relativity: The study of very fast (approaching c) objects, with no externalforces. We need to cover some of this in 311 to discuss photons, β-decay electrons,

and their kinematics. Why?

Because photons travel at the speed of light. In Nuclear Engineering applications,photons arise from

• nuclear disintegrations (fission)

• nuclear transitions (isotopes, excited states)

• electrons slowing down in materials (bremsstrahlung [Medical Physics, medicalimaging using X-rays])

• Glow Blue! That’s really Cherenkov radiation.See http://en.wikipedia.org/wiki/Cherenkov radiation


Fission


Isotopes

The 60Co decay scheme ...


X-Rays for cancer treatment

A medical LINAC used for treatments of many diseases (particularly cancer)...


X-Rays for Diagnostic Imaging

A CT scanner...


Blue Light

Pavel Alekseyevich Cherenkov (1904-1990) and his eponymous effect (ATR/INL)


1.3 Some History

(L→R): Isaac Newton (1642–1727), Joseph-Louis de Lagrange (1736–1813), and William Rowe Hamilton (1805-1865)

Around 1900 or so, all (in principle) was known!—or so it seemed, to followers of the above austere-looking gentlemen

Classical Mechanics = “The Theory of Everything” a.k.a. “Holy Grail”

Some famous scientists said:”If it has mass, and velocity, I can predict its evolution in time, forever.”

They later retracted their statements!

The motions of everything from dust particles to galaxies, were considered to beknowable, and predictable.


Maxwell, and his equations

(L→R): Michael Faraday (1791–1867), Andre-Marie Ampere (1775–1836)

James Clerk Maxwell (1831–1879) and Carl Friedrich Gauss (1777-1855)

Gauss’s Law of Electricity: ~∇ · ~D = 4πρ

Gauss’s Law of Magnetism: ~∇ · ~B = 0

Faraday’s Law: ~∇× ~E = −1

c

∂ ~B

∂t

Ampere-Maxwell Law: ~∇× ~H =4π

c~J +

1

c

∂ ~D

∂t


Telescope Observations

Newer observations of distant galaxies that move with v ≈ c, seemed to be . . . . . . . . . . . .

. . . . . . . . . . . . . . . governed by different physical laws. . . . . . . . . . . . . . . . Unacceptable!


1.4 Review of Classical Physics

1.4.1 Mechanics ...

Kinetic energy (a scalar) of a mass m moving with velocity ~v in some direction:

K =1

2mv2 . (1.1)

Momentum (a vector), is:

~p = m~v . (1.2)

(1.1) → (1.2) ⇒:

~p · ~p = p2 = ~m~v · ~m~v = m2~v · ~v = m2v2 . (1.3)

(1.1) & (1.3)/(2m) ⇒:

p2

2m=

m2v2

2m=

1

2mv2 = K .

Hence:

K =p2

2m. (1.4)


1.4.1 ...Mechanics ...

The kinetic energy change across a potential difference of 1 volt is:

∆K = 1eV ≈ 1.602× 10−19 J (1.5)

for a singly charged particle (one unit of charge e).


Review of 2-body Collision Kinematics

What is it?.........................Newtonian mechanics

Bang two things together, and two things separate after the collision

Simplification/specification:

- Two bodies in, two bodies out- No bodies are rotating, before or after the collision


Why do we study this? NE/RS applications!

neutron-neutron scatteringneutron-proton scattering (reactor cooling)neutron-deuteron scattering (CANDU reactor)proton/alpha/electron-nuclearscattering (Rutherford scattering)neutron-nuclear scattering (fuel bundles, containment)neutron shieldingfission ...


How do we study this?

Abbreviation Conservation Law Conserved quantityCoE Conservation of energy kinetic energy, and “other” things

Co~M... Conservation of momentum ~p has 3 components...CoMx ” in the x-direction ~p · x or pxCoMy ” in the y-direction ~p · y or pyCoMz ” in the z-direction ~p · z or pz

CoMmass, CoMa, CoM Conservation of mass no change in classical physics


Conservation of Energy implies (Shorthand: CoE⇒):

1

2m1(v

21,x+ v21,y+ v21,z)+

1

2m2(v

22,x+ v22,y+ v22,z)+Q =

1

2ma(v

2a,x+ v2a,y+ v2a,z)+

1

2mb(v

2b,x+ v2b,y+ v2b,z)(1.6)

Conservation of Momentum in the x-direction implies (Shorthand: CoMx ⇒):

m1v1,x +m2v2,x = mava,x +mbvb,x (1.7)

Conservation of Momentum in the other 2 spatial coordinates implies:

CoMy ⇒:

m1v1,y +m2v2,y = mava,y +mbvb,y (1.8)

CoMz ⇒:

m1v1,z +m2v2,z = mava,z +mbvb,z (1.9)

Conservation of Mass1 implies (Shorthand: CoMa⇒):

m1 +m2 = ma +mb (1.10)

What is Q?


1Aside: Later on, when we discuss relativistic kinematics, we shall see that Q = (m1 +m2 −ma −mb)c2, where c is the speed of light. The mass differences are usually so

small that Conservation of Mass is employed without loss of accuracy in a non-relativistic analysis. However, classical analysis does not make use of this directly.

Q, the “reaction” Q-value ...

Describes the type of reaction:

Q > 0 “exothermic”, “exoergic” Koutgoing > Kincoming

Q < 0 “endothermic”, “endoergic” Koutgoing < Kincoming

Q = 0 “elastic” Koutgoing = Kincoming

Examples:

Q > 0 “exothermic”, “exoergic” fissionQ < 0 “endothermic”, “endoergic” two magnets collide and stickQ = 0 “elastic” billiards (very close to zero)


Complexity!

Eqns: (1.6)–(1.10) have up to 17 unknowns: m1, ~v1, m2, ~v2, ma, ~va, mb, ~vb, and Q

In more condensed vector notation,

CoE ⇒ m1v21

2+

m2v22

2+Q =

mav2a

2+mbv

2b

2, (1.11)

Co~M ⇒ m1~v1 +m2~v2 = ma~va +mb~vb . (1.12)

If we know some of the variables, we can develop interesting relationships between them.


Solution Strategies

Here is an example demonstrating how to eliminate ~vb.

Isolate ~vb and rearrange (1.12) and (1.11) using (1.12)2 - 2mb× (1.11) to give:

(m1~v1 +m2~v2 −ma~va)2 −mb(m1v

21 +m2v

22 −mav

2a)− 2mbQ = 0 , (1.13)

vb has vanished!

Fortunately, many interesting cases are much simpler.


2D elastic collision, equal masses ...

If you play billiards, or know how a Newton’s cradle works, that helps!This is a good model for neutron-neutron scattering! It is 3D process.Why is is called 2D? (Discuss)

1. Set up the CoE and Co ~M equations assuming ~v2 = 0 and Q = 0:

1

2mv21 =

1

2mv2a +

1

2mv2b (1.15)

m~v1 = m~va +m~vb (1.16)

2. |~va + ~vb|2 = v21 is formed using |(1.16)|2/m2

v2a + v2b = v21 is formed using (1.15)/(m/2)

3. Subtract the above two equations to make v21 vanish ⇒v21 − v21 = |~va + ~vb|2 − (v2a + v2b )

✘✘✘✘✘✘✘✘✘✿0

(v21 − v21) = |~va|2 + 2~va · ~vb + |~vb|2 − (v2a + v2b)

✘✘✘✘✘✘✘✘✘✿0

(v21 − v21) =✘✘✘✘✘✘✘✘✘✘✘✿0

(|~va|2 − v2a) + 2~va · ~vb +✘✘✘✘✘✘✘✘✘✘✿0

(|~vb|2 − v2b)

⇒~va · ~vb = 0 !!!!!!!!!!!! , (1.17)


...2D elastic collision, equal masses ...

Interpretation?

Rewrite (1.17), ~va · ~vb = 0 as2:

|~va||~vb| cos Θ = vavb cosΘ = 0


2 There are some important conventions here:

Θ: angle between the projectile and the target particles, after the collision. Sometimes you see θab for this.

θ: angle between the projectile’s initial and final directions.

ϕ: angle between the target’s final direction and the projectile’s final direction. Sometimes φ is used for the ϕ symbol

Note that : Θ+ ϕ = Θ.


Conclusions, for starting condition: v1 = v0, v2 = 0 (target at rest).

We enumerate all the possibilities described by the solution vavb cosΘ = 0.

• If va 6= 0 and vb 6= 0 then cos Θ = 0 ⇒ Θ = π/2 or 90◦.θ falls in the range, 0 < θ < π/2, and ϕ is anti-correlated with it by ϕ = π/2− θ.

• After the collision, the projectile is at rest, viz. va = 0. Θ is undefined.Projectile scores a direct hit! You can show that ~va = ~v0 in this case.

It goes straight ahead!



• After the collision, the target is at rest, viz. vb = 0. Θ is undefined.Projectile missed the target! You can show that ~va = ~v0 in this case.

• The projectile and the target are at rest. v1 = v2 = va = vb = 0.Θ is undefined. Nothing is happening!

All these scenarios are described by the one equation vavb cos Θ = 0.



The math is even more general than that!Imagine that the target object is hidden.

This is always the case when dealing with nuclear radiations.There are a few things you can say:

1. the projectile came out, or

2. the target particle came out, or

3. the target and projectile exchanged matter, and something looking identical came out!

You can never be sure!3


3Oddly enough, if you suggested the third scenario to most people, they would justifiably give you a wide berth! However, in Quantum Mechanics, the answer is more

analogous to the third possibility. Particles are also waves. Their wave functions interact with each other constructively and destructively, and participate in the propagation

of a composite wavefunction. There is much more about this story that we will talk about later in the course... It gets curiouser, and curiouser!

...2D elastic collision, equal masses

Kinematic analysis is a powerful tool, and a gross over-simplification!

It does not tell anything about how the particles interact.

That requires a discussion of “dynamics” and is much more complicated.

See Wiki: http://en.wikipedia.org/wiki/Dynamics_(mechanics)

However, the conservation laws still apply, and they allow us to say many things aboutcollisions (sometimes ... surprising things!)

Moreover, these kinematical relations hold for subatomic particles, billiard balls, asteroids,planets and galaxies.

And who knows, maybe even for universes.


Example: Elastic collision, unequal masses ...

This is like the billiard ball collision, but the masses of the target and projectile are different.

With Q = 0, CoE and Co~M ⇒:

1

2mv20 =

1

2mv2 +

1

2MV 2 ; m~v0 = m~v +M~V (1.18)

We are looking for θ, therefore, we must eliminate V . Reorganize (1.18)

v20 − v2 = nV 2 (1.19)

~v0 − ~v = n~V (1.20)

where n ≡ M/m, the ratio of the target’s mass to the projectile’s mass.

n× (1.19)− |(1.20)|2 ⇒ n(v20 − v2)− (v20 − 2~v0 · ~v + v2) = 0 (1.21)

V is eliminated!


... Elastic collision, unequal masses ...

Reorganizing (1.21) gives a quadratic equation in v:

(n + 1)v2 − 2vv0 cos θ − (n− 1)v20 = 0 (1.22)

that is easily solved4 to give the relationship between v, and θ, that comes in two forms:

v

v0=

cos θ ±√n2 − 1 + cos2 θ

n + 1(1.23)

cos θ ±√

n2 − sin2 θ

n + 1(1.24)

Case 1: M > m ⇒ n > 1, target is heavierIn this case, only the + sign in (1.24) applies because v must be ≥ 0 (v is the magnitudeof a vector). Hence, the projectile can scatter forward or backward, in fact, any angle!That is 0 ≤ θ ≤ π, and therefore, 0 ≤ v ≤ v0.


4We shall encounter, quite often in this course, quadratic equations of the form ax2 + 2bx+ c = 0, that have the solutions: x = [−b±√b2 − ac]/a.

... Elastic collision, unequal masses

Case 2: M < m ⇒ n < 1, target is lighterIn this case both signs apply because the radical

√is less in magnitude than cos θ. There

are two consequences:

• there is a maximum angle of scatter viz. θmax = arcsinn.

• for each θ there are two values of v, associated with the two signs in the solution ofthe quadratic equation

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

cos(θ)

v/v

0

dashed lines: M > m target heavier, dotted lines: M < m target lighter, solid line: M = m, equal masses


Inelastic collisions ...

Imagine the scenrio described in the picture below where an incoming particle strikesa particle of equal mass, and sticks to it:

CoE and Co~M ⇒:1

2mv20 = 2× 1

2mV 2 (1.25)

m~v0 = 2×m~V (1.26)

The 2×m above accounts for the two particles being stuck together after the collision –equivalent to a single particle of mass 2m.Simplifying the above gives:

v20 = 2V 2 (1.27)

~v0 = 2~V that implies

v20 = 4V 2 (1.28)

which, of course, can not be true. We have missed something! Include Q!


Inelastic collisions with Q

If we include Q, CoE and Co~M ⇒:

1

2mv20 +Q = 2× 1

2mV 2 (1.29)

m~v0 = 2×m~V (1.30)

The 2×m above accounts for the two particles being stuck together after the collision –equivalent to a single particle of mass 2m.Simplifying the above gives:

v20 + 2Q/m = 2V 2 (1.31)

~v0 = 2~V that implies

v20 = 4V 2 (1.32)

We can solve the above 2× (1.31) = (1.32) ⇒ Q = −14mv20 = −1

2K0

Conclusions:

• Q < 0, it’s an endothermic reaction

• half the initial kinetic energy goes into the kinetic energy of the outgoing mass

• the remaining half goes into heat, deformation etc. of the particles

• in the interest of self-preservation, don’t veer for deer


Inelastic collisions, equal masses Q ≥ −K0/2 ...

We know, from the previous problem, that when equal masses are involved, the absoluteminimum Q is −K0/2. Now we examine all the other possibilities.

Special case |~v1| = |~v2| ≡ |~v|

In this case, the outgoing particles (there is no exchange of mass) have equal speed, andhence, the same kinetic energy.

CoE and Co~M ⇒:1

2mv20 +Q =

1

2mv21 +

1

2mv22 (1.33)

m~v0 = m~v1 +m~v2 (1.34)

You can show, in this case

cosΘ =−Q

K0 +Q(1.35)

Does it make sense? As Q −→ −K0/2, cosΘ −→ 1 (Previous case).

As Q −→ ∞, cosΘ −→? Explain!


... Inelastic collisions, equal masses Q ≥ −K0/2 ...

−1 0 1 2 3 4 5 6 7 8 9 10−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Q/K0

cos(Θ

)

cos(Θ) vs. Q/K0


Zero-Momentum Frame

Sometimes, it is advantageous to work in an inertial frame where there is no net momen-tum. The velocity of this frame is determined as follows:We seek a velocity ~V that, when subtracted from the velocities of the initial particles, haszero resultant momentum. Thus, referring to the left hand side (LHS) of (1.11) we want:

m1(~v1 − ~V ) +m2(~v2 − ~V ) = 0 , (1.36)

for which the solution is:

~V =m1~v1 +m2~v2m1 +m2

. (1.37)

Perform the calculation extracting the kinematic relations, in this frame. That is, ~v1 →~v1 − ~V , and so on. And then, once the desired result is obtained, transform the velocityvectors in the opposite direction.

Exercise: Repeat the case of a 2D elastic collision with equal masses in the zero-momentum

frame, transform back to the rest frame of the target particle, and show the same result

that was discussed before.


Particle-at-Rest Frame ...

At other times, it is advantageous to work in a frame where one of the particles in theinitial state is at rest (the stationary target), since in fact, this is usually the case. But,had we started from the situation described in (1.11) and (1.12) we may transform to areference frame where the target (m2, say) is at rest by subtracting ~v2 from all velocitiesin the problem, then rotating the new ~v1 into the z-direction, and the new ~va and ~vbinto the xz-plane, so it can be visualized on the blackboard, or paper. After the requiredkinematical quantities are obtained, reverse the rotations5, and reverse the velocity trans-formation and you’re done.

This situation (target-at-rest) is very important since it often represents the starting phys-ical condition.

Let us illustrate what happens to (1.13) in this circumstance.Setting ~v2 = 0 in (1.13) gives us:

m2av

2a − 2m1mav1va cos θa +m2

1v21 −mb(m1v

21 −mav

2a)− 2mbQ = 0 , (1.38)

where θa is the angle of ma wrt (“wrt” is shorthand for “with respect to”) the z-axis, inspherical-polar coordinates.


5Rotations do not commute, so be careful of the order!

... Particle-at-Rest Frame ...

Or, more suggestively,

[ma(ma +mb)]v2a + [−2m1mav1 cos θa]va + [m1(m1 −mb)v

21 − 2mbQ] = 0 . (1.39)

This can be regarded as a solution of θa in terms of va, or a quadratic formula of the form

Ax2 +Bx + C = 0 ,

for which the solution of va in terms of θa is well known. (A, B, C are constants, and weseek the values of x for which the above is true.) We shall leave further exploration of thisquadratic solution to the examples and problems. However, the quadratic equation has arich variety of solutions. Sometimes, there are no solutions (kinematically restricted) orone or two solutions. We shall revisit this topic later in the course.


... Particle-at-Rest Frame ...

Illustration

Consider a collision between two identical masses, traveling with equal but opposite ve-locities along the x-direction, that produce two masses, ma and mb.By applying the conservation laws we just discussed, we can show that the resultantvelocities are:

vb = ±√

ma

mbv ; va = ∓

√

mb

mav ,

where the final velocities are oriented in directions opposite.

This makes sense! There is no net momentum before the collision, so you expect thelighter particle to be moving faster and in the opposite direction to the heavier particle.It is also easy to verify that the initial and final kinetic energies are the same.


Potential, Kinetic, Total Energy, Angular Momentum ...

A particle (Let’s think of this as a small object.) moving in one dimension can be subjectedto a force F (x) that causes it to move. Associated with this force is a potential, U (x).External forces and potentials are related by:

F (x) = −dU (x)

dx. (1.40)

In three dimensions, we have F (~x) and U (~x), that are similarly related:

F (~x) = −~∇U (~x) . (1.41)

The total energy, E, of a particle subjected to an external force is, in one dimension, aconstant:

E = K(x) + U (x) . (1.42)


...Potential, Kinetic, Total Energy, Angular Momentum ...

In three dimensions, a system of N particles has its total energy conserved as well, andmay be written:

E =N∑

i=1

[K(~xi) + U (~xi)] . (1.43)

Angular momentum ~L for a single particle is also conserved. It is expressed in terms ofthe cross product of the position and momentum as:

~L = ~x× ~p . (1.44)

In three dimensions, a system of N particles has its total angular momentum conservedas well, and may be written:

~L =N∑

i=1

~xi × ~pi . (1.45)


1.6. Significant Figures ...

Physical things are measured two ways. There are integral quantities, e.g. 3 electrons, 4photons, etc. It doesn’t make sense to say “A tritium nucleus is made up of 1.0 protonsand 2.0 neutrons.”, because 1.0 and 2.0 are not integers, and there should be no ambiguityin the number of discrete particles. The correct statement would be, “A tritium nucleusis made up of 1 proton and 2 neutrons.”

Other quantities are not discrete. There is ambiguity in the statement, ”The α-particlehas an energy of 5 MeV.”, because there is no statement as to how good the “5” is. Thebest we can do, is to give to a precision, the precision of that participant, with the leastprecision. That’s not the best situation, but it is what we have to live with, without moreinformation.

For example, consider:

r = a � b ,

where � represents one of the operators, ×, ÷, +, or −. Define a “precision operator”,P(), that returns the precision of its argument. For example, P(1.37) = P(0.000137) = 3.Then,

P(r) = minP(a),P(b) .


... Significant Figures

If you come across a statement like, ”The α-particle has an energy of 5 MeV.”, it is likelyan error, because no educator wishes to be ambiguous. If he or she does, and it’s onpurpose, then the question was likely designed to make you think. It is valid to say, ”Theα-particle has an energy of 5 MeV (exactly).”, in which case, you are expected to treatthe “5” as an exact real number, with infinite precision.

In some situations, you will be told, ”The α-particle has an energy of 5.02±0.13 MeV.”,in which case you will be expected to estimate the error on the final quantity, if you useit in a computation. This is discussed later in this chapter.


†1.8. Basic Error Estimation ...

Accounting for Estimated Error for Independent Quantities

Suppose M is a measured quantity, the result of some calculation or experiment, thatdepends on two other quantities, a and b, that each have their own estimated uncertain-ties σa and σb. Assuming that the relative errors in a and b are small (i.e. we propagatefirst-order errors only), and a and b are independent (the result of two independent mea-surements), then the estimated error in M , namely σM , is determined as follows:

σM =

√

(

∂M(a, b)

∂a

)2

σ2a +

(

∂M(a, b)

∂b

)2

σ2b , (1.46)

and the fractional error is:

σMM

=1

M(a, b)

√

(

∂M(a, b)

∂a

)2

σ2a +

(

∂M(a, b)

∂b

)2

σ2b . (1.47)

The extension to measurements that have three or more independent inputs, is straight-forward.


... Basic Error Estimation ...

Example 1

Question: After a collision reaction involving two protons, the two protons have kineticenergies 2.8±0.6 and 10.0±1.0 MeV. What is the total combined resultant kinetic energyand its estimated error?

Answer: The total energy is

K = K1 +K2 .

Therefore, by the rule stated above,

σK =√

σ2K1

+ σ2K2

.

Therefore,

K = 12.8± 1.2 MeV .

Food for thought: How should you state the result if the energy of the second proton was10±1 MeV?


... Basic Error Estimation ...

Example 2

Question: M(a, b, c) = ab2c3. What is the relative estimated error in M?

Answer: You can show that

σMM

=

√

(σaa

)2

+ 4(σbb

)2

+ 9(σcc

)2

.

Note: Don’t expect this very neat expression for all calculations of relative error. It isreally a fortuitous result of the given form of M .


Documents

1. Introduction - University of Michiganners312/CourseLibrary/Lecture01.pdf · 1Aside: Later on, when we discuss relativistic kinematics, we shall see that Q = (m1 +m2 −m a −m